Problem: Integrate. $ \int 10\sin(x)\,dx $ $=$ $+ C$
Solution: We need a function whose derivative is $10\sin(x)$. We know that the derivative of $\cos(x)$ is $-\sin(x)$, so let's start there: $\dfrac{d}{dx} \cos(x) = -\sin(x)$ Now let's add in a negative sign: $\dfrac{d}{dx}\left[ -\cos(x)\right] = \sin(x)$ Now let's multiply by $10$ : $\dfrac{d}{dx}\left[ -10\cos(x) \right]= 10\sin(x)$ Because finding the integral is the opposite of taking the derivative, this means that: $ \int 10\sin(x)\,dx =-10 \cos(x)\, + C$ The answer: $-10 \cos(x)\, + C$